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  Leetcode-424-替换后的最长重复字符
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      <h1 id="Leetcode-424-替换后的最长重复字符"><a href="#Leetcode-424-替换后的最长重复字符" class="headerlink" title="Leetcode-424-替换后的最长重复字符"></a>Leetcode-424-<a href="https://leetcode-cn.com/problems/longest-repeating-character-replacement/" target="_blank" rel="noopener">替换后的最长重复字符</a></h1><h2 id="题目描述"><a href="#题目描述" class="headerlink" title="题目描述"></a>题目描述</h2><p>给你一个仅由大写英文字母组成的字符串，你可以将任意位置上的字符替换成另外的字符，总共可最多替换 k 次。在执行上述操作后，找到包含重复字母的最长子串的长度。</p>
<p>注意：字符串长度 和 k 不会超过 104。</p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br></pre></td><td class="code"><pre><span class="line">示例 1：</span><br><span class="line"></span><br><span class="line">输入：s &#x3D; &quot;ABAB&quot;, k &#x3D; 2</span><br><span class="line">输出：4</span><br><span class="line">解释：用两个&#39;A&#39;替换为两个&#39;B&#39;,反之亦然。</span><br><span class="line"></span><br><span class="line">示例 2：</span><br><span class="line"></span><br><span class="line">输入：s &#x3D; &quot;AABABBA&quot;, k &#x3D; 1</span><br><span class="line">输出：4</span><br><span class="line">解释：</span><br><span class="line">将中间的一个&#39;A&#39;替换为&#39;B&#39;,字符串变为 &quot;AABBBBA&quot;。</span><br><span class="line">子串 &quot;BBBB&quot; 有最长重复字母, 答案为 4。</span><br></pre></td></tr></table></figure>



<a id="more"></a>



<h2 id="方法一-：-滑动窗口"><a href="#方法一-：-滑动窗口" class="headerlink" title="方法一 ： 滑动窗口"></a>方法一 ： 滑动窗口</h2><p><strong>说一下暴力解法：</strong></p>
<ul>
<li><p>如果一个问题暂时没有思路，可以先考虑暴力解法（不一定要实现）。<strong>当前问题的暴力解法是：枚举输入字符串的 所有 子串，对于每一个子串：</strong></p>
<ul>
<li>如果子串里所有的字符都一样，就考虑长度更长的子串；</li>
<li>如果当前子串里出现了至少两种字符，要想使得替换以后所有的字符都一样，并且重复的、连续的部分更长，<strong>应该替换掉出现次数最多字符 以外 的字符。</strong></li>
</ul>
</li>
</ul>
<p><strong>暴力解法的缺点：</strong></p>
<ul>
<li>做了重复的工作，子串和子串有很多重合的部分，重复扫描它们是不划算的；</li>
<li>做了很多没有必要的工作：<ul>
<li><strong>如果找到了一个长度为 L 且替换 k 个字符以后全部相等的子串</strong>，就没有必要考虑长度小于等于 L 的子串，因为题目只让我们找到符合题意的最长的长度；</li>
<li><strong>如果找到了一个长度为 L 且替换 k 个字符以后不能全部相等的子串</strong>，左边界相同、长度更长的子串一定不符合要求（原因我们放在最后说）。</li>
</ul>
</li>
</ul>
<hr>
<p><strong>滑动窗口</strong></p>
<ul>
<li>以 <code>s = AABCABBB</code>，<code>k = 2</code> 为例，寻找替换 <code>k</code> 次以后字符全部相等的最长子串的长度的过程如下图所示：</li>
</ul>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20210204/104646614.png" alt="mark"></p>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20210204/104658660.png" alt="mark"></p>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20210204/104706538.png" alt="mark"></p>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20210204/104713644.png" alt="mark"></p>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20210204/104722926.png" alt="mark"></p>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20210204/104731243.png" alt="mark"></p>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20210204/104740342.png" alt="mark"></p>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20210204/112527361.png" alt="mark"></p>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20210204/113537783.png" alt="mark"></p>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20210204/113631793.png" alt="mark"></p>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20210204/113644299.png" alt="mark"></p>
<hr>
<ul>
<li><p>整个过程，我们使用了两个表示边界的变量，一前一后，交替在字符串上前进</p>
</li>
<li><p>右边界先向右移动，直到不能移动了为止，左边界再继续向右移动，整个过程像极了一条滑动的窗口在线段上移动。</p>
</li>
<li><p>除此之外，考虑的子串中最多出现的字符是次数，因此须要一个频数数组，记录每个字符出现的次数。</p>
</li>
</ul>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20210204/114022723.png" alt="mark"></p>
<figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span> </span>&#123;</span><br><span class="line">    <span class="function"><span class="keyword">public</span> <span class="keyword">int</span> <span class="title">characterReplacement</span><span class="params">(String s, <span class="keyword">int</span> k)</span> </span>&#123;</span><br><span class="line">        <span class="comment">// 1. 特判</span></span><br><span class="line">        <span class="keyword">int</span> len = s.length();</span><br><span class="line">        <span class="keyword">if</span>(len &lt; <span class="number">2</span>)&#123;                                <span class="comment">// 长度为0或1</span></span><br><span class="line">            <span class="keyword">return</span> len;</span><br><span class="line">        &#125;</span><br><span class="line"></span><br><span class="line">        <span class="comment">// 2. 初始化条件</span></span><br><span class="line">        <span class="keyword">char</span>[] charArray = s.toCharArray();</span><br><span class="line">        <span class="keyword">int</span> left  = <span class="number">0</span>;</span><br><span class="line">        <span class="keyword">int</span> right = <span class="number">0</span>;</span><br><span class="line"></span><br><span class="line">        <span class="keyword">int</span> res = <span class="number">0</span>;</span><br><span class="line">        <span class="keyword">int</span> maxCount = <span class="number">0</span>;</span><br><span class="line">        <span class="keyword">int</span>[] freq = <span class="keyword">new</span> <span class="keyword">int</span>[<span class="number">26</span>];                   <span class="comment">// 记录字符出现的频率</span></span><br><span class="line"></span><br><span class="line">        <span class="comment">// 3. 遍历一次字符数组</span></span><br><span class="line">        <span class="keyword">while</span>(right &lt; len)&#123;</span><br><span class="line">            <span class="comment">// 3.1 在这里维护maxCount,因为每一次右边界读入一个字符，字符的频率增加，才会使得maxCount增加</span></span><br><span class="line">            freq[charArray[right] - <span class="string">'A'</span>]++;</span><br><span class="line">            maxCount = Math.max(maxCount,freq[charArray[right] - <span class="string">'A'</span>]);</span><br><span class="line">            right++;</span><br><span class="line"></span><br><span class="line">            <span class="comment">// 3.2 </span></span><br><span class="line">            <span class="keyword">if</span>(right - left &gt; maxCount + k)&#123;        <span class="comment">// 说明此时的k不够用，其他不适 最多出现的字符替换以后，都不能填满这个滑动窗口</span></span><br><span class="line">                freq[charArray[left] - <span class="string">'A'</span>]--;      <span class="comment">// 移出滑动窗口的时候，频数数组须要相应地做减法</span></span><br><span class="line">                left++;                             <span class="comment">// 左边界向右移动</span></span><br><span class="line">            &#125;</span><br><span class="line">            <span class="comment">// 3.3 </span></span><br><span class="line">            res = Math.max(res,right - left);</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> res;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>

<p><strong>复杂度分析：</strong></p>
<ul>
<li>时间复杂度：O(N)，这里 N 是输入字符串 <code>S</code> 的长度；</li>
<li>空间复杂度：O(A)，这里 A<em>A</em> 是输入字符串 <code>S</code> 出现的字符 ASCII 值的范围</li>
</ul>

      
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